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inessss [21]
2 years ago
13

What is the solution to the inequality -3x - 42 > 3

Mathematics
1 answer:
rosijanka [135]2 years ago
5 0

Answer:

I don’t know which answer choice it is since C and D are the same but the answer is supposed to be x < -15.

Step-by-step explanation:

the image has the work shown

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−1/2 (x+2) + 112x = 3<br><br> x=?
AysviL [449]

Given equation :−1/2 (x+2) + 112x = 3

We have -1/2 in front of parenthesis on left side .

It's better to remove fraction in an equation, in order make it easier to solve.

In order to remove 2 from denominator of 2, we need to multiply each and every term by 2.

Multiplying each term in the equation by 2, we get

2* −1/2(x+2) +2* 112x = 2*3.

On simplifying this step, we get

-1(x+2) +224x = 6.

Distributing -1 over (x+2), we get

-x -2 +224x = 6

Combining like terms on left side, -x+224x=223x

223x -2 = 6

Adding 2 on both sides, we get

223x -2+2 = 6+2

223x = 8

Dividing both sides by 223, we get

223x/223 = 8/223.

x= 8/223.

6 0
3 years ago
How to solve y in -(y+9)=10
Allisa [31]

Answer:


Step-by-step explanation:

- (y+9) = 10


Multiply through by -1 gives us


y+9 = -10


Subtract 9 from both sides


y = -19


That's your answer.


If you work it through it satisfies the equation.


That is:


- (-19 +9) = 10


-(-10) = 10


10 = 10


Boo ya!

5 0
3 years ago
What is 11mp in expanded form
Flauer [41]

Answer:

<h3>Eleven-per miles who doesn't know that or 11.✌️</h3>
4 0
3 years ago
3. Why does it tako 3 copies of; to show the same amount as 1 copy of ?? Explain your answer in words
Amanda [17]
Because 3/6 is equal to 1/2.
You multiply the 1 (in 1/2) by 3, and the 2 (in 1/2) by 3 to get 3/6. You need to have a common denominator, and 6 was the common denominator.
7 0
2 years ago
Help me with differentation and integration please!!
Marina86 [1]

Answer:

See below

Step-by-step explanation:

\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x

Recall

\dfrac{d}{dx}\tan x=\sec^2

Using the chain rule

\dfrac{dy}{dx}= \dfrac{dy}{du} \dfrac{du}{dx}

such that u = \tan x

we can get a general formulation for

y = \tan^n x

Considering the power rule

\boxed{\dfrac{d}{dx} x^n = nx^{n-1}}

we have

\dfrac{dy}{dx} =n u^{n-1} \sec^2 x \implies \dfrac{dy}{dx} =n \tan^{n-1} \sec^2 x

therefore,

\dfrac{d}{dx}\tan^3 x=3\tan^2x \sec^2x

Now, once

\sec^2 x - 1= \tan^2x

we have

3\tan^2x \sec^2x =  3(\sec^2 x - 1) \sec^2x = 3\sec^4x-3\sec^2x

Hence, we showed

\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x

================================================

For the integration,

$\int \sec^4 x\, dx $

considering the previous part, we will use the identity

\boxed{\sec^2 x - 1= \tan^2x}

thus

$\int\sec^4x\,dx=\int \sec^2 x(\tan^2x+1)\,dx = \int \sec^2 x \tan^2x+\sec^2 x\,dx$

and

$\int \sec^2 x \tan^2x+\sec^2 x\,dx = \int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx $

Considering u = \tan x

and then du=\sec^2x\ dx

we have

$\int u^2 \, du = \dfrac{u^3}{3}+C$

Therefore,

$\int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx = \dfrac{\tan^3 x}{3}+\tan x + C$

$\boxed{\int \sec^4 x\, dx  = \dfrac{\tan^3 x}{3}+\tan x + C }$

6 0
2 years ago
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